cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 17 results. Next

A383057 Decimal expansion of the asymptotic mean of A056671(k)/A034444(k), the ratio between the number of squarefree unitary divisors and the number of unitary divisors over the positive integers.

Original entry on oeis.org

7, 8, 9, 3, 6, 2, 6, 0, 1, 2, 6, 0, 9, 8, 9, 0, 2, 9, 1, 0, 3, 7, 0, 8, 6, 2, 9, 2, 5, 1, 3, 9, 6, 8, 9, 2, 7, 6, 8, 5, 1, 6, 7, 6, 0, 5, 2, 6, 9, 1, 6, 5, 0, 5, 3, 3, 3, 6, 8, 4, 7, 4, 1, 6, 1, 3, 6, 0, 9, 9, 3, 9, 8, 8, 2, 2, 5, 2, 7, 5, 3, 6, 3, 2, 5, 0, 2, 0, 3, 4, 3, 4, 4, 8, 7, 0, 9, 9, 0, 8, 4, 9, 1, 1, 4
Offset: 0

Views

Author

Amiram Eldar, Apr 15 2025

Keywords

Comments

The asymptotic mean of the inverse ratio A034444(k)/A056671(k) is 15/Pi^2 (A082020).

Examples

			0.78936260126098902910370862925139689276851676052691...
		

Crossrefs

The unitary analog of A308043.

Programs

  • Mathematica
    $MaxExtraPrecision = 300; m = 300; f[p_] := 1 - 1/(2*p^2); c = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x]]; RealDigits[Exp[NSum[Indexed[c, n]*(PrimeZetaP[n]), {n, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]
  • PARI
    prodeulerrat(1 - 1/(2*p^2))

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A056671(k)/A034444(k).
Equals Product_{p prime} (1 - 1/(2*p^2)).

A055231 Powerfree part of n: product of primes that divide n only once.

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 1, 1, 10, 11, 3, 13, 14, 15, 1, 17, 2, 19, 5, 21, 22, 23, 3, 1, 26, 1, 7, 29, 30, 31, 1, 33, 34, 35, 1, 37, 38, 39, 5, 41, 42, 43, 11, 5, 46, 47, 3, 1, 2, 51, 13, 53, 2, 55, 7, 57, 58, 59, 15, 61, 62, 7, 1, 65, 66, 67, 17, 69, 70, 71, 1, 73, 74, 3, 19, 77, 78, 79, 5
Offset: 1

Views

Author

Labos Elemer, Jun 21 2000

Keywords

Comments

The previous name was: Write n = K^2*F where F is squarefree and F = g*f where g = gcd(K,F) and f = F/g; then a(n) = f(n) = F(n)/g(n). Thus gcd(K^2,f) = 1.
Differs from A007913; they coincide if and only if g(n) = 1.
a(n) is the powerfree part of n; i.e., if n=Product(pi^ei) over all i (prime factorization) then a(n)=Product(pi^ei) over those i with ei=1; if n=b*c^2*d^3 then a(n) is minimum possible value of b. - Henry Bottomley, Sep 01 2000
Also denominator of n/rad(n)^2, where rad is the squarefree kernel of n (A007947), numerator: A062378. - Reinhard Zumkeller, Dec 10 2002
Largest unitary squarefree number dividing n (the unitary squarefree kernel of n). - Steven Finch, Mar 01 2004
From Bernard Schott, Dec 19 2022: (Start)
a(n) = 1 iff n is a squareful number (A001694).
1 < a(n) < n iff n is a nonsquarefree number that is not squareful (A332785).
a(n) = n iff n is a squarefree number (A005117). (End)

Crossrefs

Positions of 1's: A001694.
Cf. A008833, A007913, A007947, A000188, A057521, A055773 (computed for n!), A056169 (number of prime divisors), A056671 (number of divisors), A092261 (sum of divisors of the n-th term), A197863, A332785.
Cf. A005117 (subsequence).

Programs

  • Maple
    A055231 := proc(n)
        a := 1 ;
        if n > 1 then
            for f in ifactors(n)[2] do
                if op(2, f) = 1 then
                    a := a*op(1, f) ;
                end if;
            end do:
        end if;
        a ;
    end proc: # R. J. Mathar, Dec 23 2011
  • Mathematica
    rad[n_] := Times @@ First /@ FactorInteger[n]; a[n_] := Denominator[n/rad[n]^2]; Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Jun 20 2013, after Reinhard Zumkeller *)
    f[p_, e_] := If[e==1, p, 1]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *)
  • PARI
    A055231(n)={
       local(a=1);
       f=factor(n) ;
       for(i=1,matsize(f)[1],
             if( f[i,2] ==1, a *=  f[i,1]
             )
       ) ;
       a ;
    } /* R. J. Mathar, Mar 12 2012 */
    
  • PARI
    a(n) = {my(f=factor(n)); for (k=1, #f~, if (f[k,2] > 1, f[k,2] = 0);); factorback(f);} \\ Michel Marcus, Aug 27 2017
    
  • Python
    from math import prod
    from sympy import factorint
    def A055231(n): return prod(p for p, e in factorint(n).items() if e == 1) # Chai Wah Wu, Nov 14 2022
  • Scheme
    ;; With memoization-macro definec.
    (definec (A055231 n) (if (= 1 n) 1 (* (if (= 1 (A067029 n)) (A020639 n) 1) (A055231 (A028234 n))))) ;; Antti Karttunen, Nov 28 2017
    

Formula

a(n) = A007913(n)/gcd(A008833(n), A007913(n)).
a(n) = n/A057521(n).
Multiplicative with a(p) = p and a(p^e) = 1 for e > 1. - Vladeta Jovovic, Nov 01 2001
Dirichlet g.f.: zeta(s)*Product_{primes p} (1 + p^(1-s) - p^(-s) - p^(1-2s) + p^(-2s)). - R. J. Mathar, Dec 21 2011
a(n) = A007947(n)/A071773(n). - observed by Velin Yanev, Aug 27 2017, confirmed by Antti Karttunen, Nov 28 2017
a(1) = 1; for n > 1, a(n) = A020639(n)^A063524(A067029(n)) * a(A028234(n)). - Antti Karttunen, Nov 28 2017
a(n*m) = a(n)*a(m)/(gcd(n,a(m))*gcd(m,a(n))) for all n and m > 0 (conjectured). - Velin Yanev, Feb 06 2019. [This follows easily from the comment of Vladeta Jovovic. - N. J. A. Sloane, Mar 14 2019]
From Vaclav Kotesovec, Dec 19 2019: (Start)
Dirichlet g.f.: zeta(s-1) * zeta(s) * Product_{primes p} (1 - p^(1-3*s) + p^(2-3*s) - p^(2-2*s) + p^(-2*s) - p^(-s)).
Sum_{k=1..n} a(k) ~ c * Pi^2 * n^2 / 12, where c = Product_{primes p} (1 - 2/p^2 + 2/p^4 - 1/p^5) = 0.394913518073109872954607634745304266741971541072... (End)
a(n) = A197863(n)/n. - Amiram Eldar, Sep 01 2023

Extensions

Name replaced with a simpler description (based on Henry Bottomley's comment) by Antti Karttunen, Nov 28 2017
Incorrect comments and example deleted by Peter Munn, Nov 30 2022

A065464 Decimal expansion of Product_{p prime} (1 - (2*p-1)/p^3).

Original entry on oeis.org

4, 2, 8, 2, 4, 9, 5, 0, 5, 6, 7, 7, 0, 9, 4, 4, 4, 0, 2, 1, 8, 7, 6, 5, 7, 0, 7, 5, 8, 1, 8, 2, 3, 5, 4, 6, 1, 2, 1, 2, 9, 8, 5, 1, 3, 3, 5, 5, 9, 3, 6, 1, 4, 4, 0, 3, 1, 9, 0, 1, 3, 7, 9, 5, 3, 2, 1, 2, 3, 0, 5, 2, 1, 6, 1, 0, 8, 3, 0, 4, 4, 1, 0, 5, 3, 4, 8, 5, 1, 4, 5, 2, 4, 6, 8, 0, 6, 8, 5, 5, 4, 8, 0, 7, 5, 7
Offset: 0

Views

Author

N. J. A. Sloane, Nov 19 2001

Keywords

Comments

Sum_{n <= x} A189021(n) ~ kx, where k is this constant. - Charles R Greathouse IV, Jan 24 2018
The probability that a number chosen at random is squarefree and coprime to another randomly chosen random (see Schroeder, 2009). - Amiram Eldar, May 23 2020, corrected Aug 04 2020

Examples

			0.428249505677094440218765707581823546...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.5.1, p. 110.
  • Manfred Schroeder, Number Theory in Science and Communication, 5th edition, Springer, 2009, page 59.

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 800; digits = 98; terms = 2000; LR = Join[{0, 0}, LinearRecurrence[{-2, 0, 1}, {-2, 3, -6}, terms+10]]; r[n_Integer] := LR[[n]]; (6/Pi^2)*Exp[NSum[r[n]*(PrimeZetaP[n-1]/(n-1)), {n, 3, terms}, NSumTerms -> terms, WorkingPrecision -> digits+10, Method -> "AlternatingSigns"]] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Apr 16 2016 *)
  • PARI
    prodeulerrat(1 - (2*p-1)/p^3) \\ Amiram Eldar, Mar 12 2021

Formula

Equals A065463 divided by A013661. - R. J. Mathar, Mar 22 2011
Equals A065473 divided by A065480. - R. J. Mathar, May 02 2019
Equals (6/Pi^2)^2 * Product_{p prime} (1 + 1/(p^3 + p^2 - p - 1)) = 1.1587609... * (6/Pi^2)^2. - Amiram Eldar, May 23 2020
Equals lim_{m->oo} (1/m) * Sum_{k==1..m} (phi(k)/k)^2, where phi is the Euler totient function (A000010). - Amiram Eldar, Mar 12 2021

Extensions

More digits from Vaclav Kotesovec, Dec 18 2019

A322483 The number of semi-unitary divisors of n.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 3, 2, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 4, 4, 2, 6, 2, 4, 3, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 6, 2, 8, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 6, 4, 6, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 6, 2, 4, 4, 4, 4, 8, 2, 6, 3, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Amiram Eldar, Dec 11 2018

Keywords

Comments

The notion of semi-unitary divisor was introduced by Chidambaraswamy in 1967.
A semi-unitary divisor of n is defined as the largest divisor d of n such that the largest divisor of d that is a unitary divisor of n/d is 1. In terms of the relation defined in A322482, d is the largest divisor of n such that T(d, n/d) = 1 (the largest divisor d that is semiprime to n/d).
The number of divisors of n that are exponentially odd numbers (A268335). - Amiram Eldar, Sep 08 2023

Examples

			The semi-unitary divisors of 8 are 1, 2, 8 (4 is not semi-unitary divisor since the largest divisor of 4 that is a unitary divisor of 8/4 = 2 is 2 > 1), and their number is 3, thus a(8) = 3.
		

References

  • J. Chidambaraswamy, Sum functions of unitary and semi-unitary divisors, J. Indian Math. Soc., Vol. 31 (1967), pp. 117-126.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Floor[(e+3)/2]; sud[n_] := If[n==1, 1, Times @@ (f @@@ FactorInteger[n])]; Array[sud, 100]
  • PARI
    a(n) = {my(f = factor(n)); for (k=1, #f~, f[k,1] = (f[k,2]+3)\2; f[k,2] = 1;); factorback(f);} \\ Michel Marcus, Dec 14 2018
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/(1-X) * 1/(1-X^2) * (1 + X - X^2))[n], ", ")) \\ Vaclav Kotesovec, Sep 06 2023

Formula

Multiplicative with a(p^e) = floor((e+3)/2).
a(n) <= A000005(n) with equality if and only if n is squarefree (A005117).
a(n) = Sum_{d|n} mu(d/gcd(d, n/d))^2. - Ilya Gutkovskiy, Feb 21 2020
a(n) = A000005(A019554(n)) (the number of divisors of the smallest number whose square is divisible by n). - Amiram Eldar, Sep 02 2023
From Vaclav Kotesovec, Sep 06 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s) * Product_{p prime} (1 + 1/p^s - 1/p^(2*s)).
Dirichlet g.f.: zeta(s)^2 * zeta(2*s) * Product_{p prime} (1 - 2/p^(2*s) + 1/p^(3*s)).
Let f(s) = Product_{p prime} (1 - 2/p^(2*s) + 1/p^(3*s)).
Sum_{k=1..n} a(k) ~ Pi^2 * f(1) * n / 6 * (log(n) + 2*gamma - 1 + 12*zeta'(2)/Pi^2 + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 2/p^2 + 1/p^3) = A065464 = 0.42824950567709444...,
f'(1) = f(1) * Sum_{p prime} (4*p-3) * log(p) / (p^3 - 2*p + 1) = 0.808661108949590913395... and gamma is the Euler-Mascheroni constant A001620. (End)

A092261 Sum of unitary, squarefree divisors of n, including 1.

Original entry on oeis.org

1, 3, 4, 1, 6, 12, 8, 1, 1, 18, 12, 4, 14, 24, 24, 1, 18, 3, 20, 6, 32, 36, 24, 4, 1, 42, 1, 8, 30, 72, 32, 1, 48, 54, 48, 1, 38, 60, 56, 6, 42, 96, 44, 12, 6, 72, 48, 4, 1, 3, 72, 14, 54, 3, 72, 8, 80, 90, 60, 24, 62, 96, 8, 1, 84, 144, 68, 18, 96, 144, 72, 1, 74, 114, 4, 20, 96, 168, 80
Offset: 1

Views

Author

Steven Finch, Feb 20 2004

Keywords

Comments

Unitary convolution of the sequence of n*mu^2(n) (absolute values of A055615) and A000012. - R. J. Mathar, May 30 2011

Crossrefs

Programs

  • Mathematica
    Table[Plus @@ Select[Divisors@ n, Max @@ Last /@ FactorInteger@ # == 1 && GCD[#, n/#] == 1 &], {n, 1, 79}] (* Michael De Vlieger, Mar 08 2015 *)
    f[p_, e_] := If[e==1, p+1, 1]; a[1]=1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 79] (* Amiram Eldar, Mar 01 2019 *)
  • PARI
    a(n) = sumdiv(n, d, d*issquarefree(d)*(gcd(d, n/d) == 1)); \\ Michel Marcus, Mar 06 2015
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 + p^2*X^3 - p*X^2 - p^2*X^2)/(1-X)/(1-p*X))[n], ", ")) \\ Vaclav Kotesovec, Aug 20 2021
  • Scheme
    ;; This implementation utilizes the memoization-macro definec for which an implementation is available at http://oeis.org/wiki/Memoization#Scheme
    ;; The other functions, A020639, A067029 and A028234 can be found under the respective entries, and should likewise defined with definec:
    (definec (A092261 n) (if (= 1 n) 1 (* (+ 1 (if (> (A067029 n) 1) 0 (A020639 n))) (A092261 (A028234 n))))) ;; Antti Karttunen, Nov 25 2017
    

Formula

Multiplicative with a(p) = p+1 and a(p^e) = 1 for e > 1. - Vladeta Jovovic, Feb 22 2004
From Álvar Ibeas, Mar 06 2015: (Start)
a(n) = a(A055231(n)) = A000203(A055231(n)).
Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + p^(1-s) - p^(1-2s)).
(End)
From Antti Karttunen, Nov 25 2017: (Start)
a(n) = A048250(A055231(n)).
a(n) = A000203(n) / A295294(n).
a(n) = A048250(n) / A295295(n) = A048250(n) / A048250(A057521(n)), where A057521(n) = A064549(A003557(n)).
(End)
Lim_{n->oo} (1/n) * Sum_{k=1..n} a(k)/k = Product_{p prime}(1 - 1/(p^2*(p+1))) = 0.881513... (A065465). - Amiram Eldar, Jun 10 2020
Dirichlet g.f.: zeta(s) * zeta(s-1) * Product_{p prime} (1 + p^(2-3*s) - p^(1-2*s) - p^(2-2*s)). - Vaclav Kotesovec, Aug 20 2021
a(n) = Sum_{d|n, gcd(d,n/d)=1} d * mu(d)^2. - Wesley Ivan Hurt, May 26 2023

A365498 Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s)).

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 1, 2, 4, 2, 4, 2, 4, 4, 1, 2, 4, 2, 4, 4, 4, 2, 2, 2, 4, 1, 4, 2, 8, 2, 1, 4, 4, 4, 4, 2, 4, 4, 2, 2, 8, 2, 4, 4, 4, 2, 2, 2, 4, 4, 4, 2, 2, 4, 2, 4, 4, 2, 8, 2, 4, 4, 1, 4, 8, 2, 4, 4, 8, 2, 2, 2, 4, 4, 4, 4, 8, 2, 2, 1, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Vaclav Kotesovec, Sep 06 2023

Keywords

Comments

The number of unitary divisors of n that are cubefree numbers (A004709). - Amiram Eldar, Sep 06 2023

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e <= 2, 2, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 06 2023 *)
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/(1-X) * (1 + X - X^3))[n], ", "))

Formula

Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Sum_{k=1..n} a(k) ~ f(1) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.5358961538283379998085026313185459506482223745141452711510108346133288...,
f'(1) = f(1) * Sum_{p prime} (-4 + 3*p + 2*p^2) * log(p) / (1 - p - p^2 + p^4) = f(1) * 1.4525924794451595590371439593828547341482465114411929136723476679...
and gamma is the Euler-Mascheroni constant A001620.
Multiplicative with a(p^e) = 2 if e <= 2, and 1 otherwise. - Amiram Eldar, Sep 06 2023
From Vaclav Kotesovec, Jan 27 2025: (Start)
Following formulas have been conjectured for this sequence by Sequence Machine, with each one giving the first 1000000 terms correctly:
a(n) = A056671(n) * A368885(n).
a(n) = A034444(n) / A368248(n).
a(n) = A158522(n) / A307428(n).
a(n) = A369310(n) / A190867(n).
a(n) = A286324(n) / A368172(n). (End)

A343443 If n = Product (p_j^k_j) then a(n) = Product (k_j + 2), with a(1) = 1.

Original entry on oeis.org

1, 3, 3, 4, 3, 9, 3, 5, 4, 9, 3, 12, 3, 9, 9, 6, 3, 12, 3, 12, 9, 9, 3, 15, 4, 9, 5, 12, 3, 27, 3, 7, 9, 9, 9, 16, 3, 9, 9, 15, 3, 27, 3, 12, 12, 9, 3, 18, 4, 12, 9, 12, 3, 15, 9, 15, 9, 9, 3, 36, 3, 9, 12, 8, 9, 27, 3, 12, 9, 27, 3, 20, 3, 9, 12, 12, 9, 27, 3, 18
Offset: 1

Views

Author

Ilya Gutkovskiy, Apr 15 2021

Keywords

Comments

Inverse Moebius transform of A056671.
a(n) depends only on the prime signature of n (see formulas). - Bernard Schott, May 03 2021

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := Times @@ ((#[[2]] + 2) & /@ FactorInteger[n]); Table[a[n], {n, 80}]
    a[n_] := Sum[If[GCD[d, n/d] == 1, DivisorSigma[0, d], 0], {d, Divisors[n]}]; Table[a[n], {n, 80}]
  • PARI
    a(n) = sumdiv(n, d, if(gcd(d, n/d)==1, numdiv(d))) \\ Andrew Howroyd, Apr 15 2021
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 + X - X^2)/(1-X)^2)[n], ", ")) \\ Vaclav Kotesovec, Feb 11 2023
    
  • Python
    from math import prod
    from sympy import factorint
    def A343443(n): return prod(e+2 for e in factorint(n).values()) # Chai Wah Wu, Feb 21 2025

Formula

a(n) = 2^omega(n) * tau_3(n) / tau(n), where omega = A001221, tau = A000005 and tau_3 = A007425.
a(n) = Sum_{d|n, gcd(d, n/d) = 1} tau(d).
From Bernard Schott, May 03 2021: (Start)
a(p^k) = k+2 for p prime, or signature [k].
a(A006881(n)) = 9 for signature [1, 1].
a(A054753(n)) = 12 for signature [2, 1].
a(A065036(n)) = 15 for signature [3, 1].
a(A085986(n)) = 16 for signature [2, 2].
a(A178739(n)) = 18 for signature [4, 1].
a(A143610(n)) = 20 for signature [3, 2].
a(A007304(n)) = 27 for signature [1, 1, 1]. (End)
Dirichlet g.f.: zeta(s)^2 * Product_{primes p} (1 + 1/p^s - 1/p^(2*s)). - Vaclav Kotesovec, Feb 11 2023
From Amiram Eldar, Sep 01 2023: (Start)
a(n) = A000005(A064549(n)).
a(n) = A363194(A348018(n)). (End)

A365499 Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^s - 1/p^(4*s)).

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 1, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 1, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 2, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 1, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 2, 1, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Vaclav Kotesovec, Sep 06 2023

Keywords

Comments

The number of unitary divisors of n that are biquadratefree numbers (A046100). - Amiram Eldar, Sep 06 2023

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e <= 3, 2, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 06 2023 *)
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/(1-X) * (1 + X - X^4))[n], ", "))

Formula

Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(4*s) + 1/p^(5*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(4*s) + 1/p^(5*s)).
Sum_{k=1..n} a(k) ~ f(1) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^4 + 1/p^5) = 0.576152735385667059520611078264117275406247116802896188543250284595724...,
f'(1) = f(1) * Sum_{p prime} (-5 + 4*p + 2*p^3) * log(p) / (1 - p - p^3 + p^5) = f(1) * 1.30114343965598023783147826007476613992233856698399986804189962...
and gamma is the Euler-Mascheroni constant A001620.
Multiplicative with a(p^e) = 2 if e <= 3, and 1 otherwise. - Amiram Eldar, Sep 06 2023

A357669 a(n) is the number of divisors of the powerful part of n.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 4, 3, 1, 1, 3, 1, 1, 1, 5, 1, 3, 1, 3, 1, 1, 1, 4, 3, 1, 4, 3, 1, 1, 1, 6, 1, 1, 1, 9, 1, 1, 1, 4, 1, 1, 1, 3, 3, 1, 1, 5, 3, 3, 1, 3, 1, 4, 1, 4, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 1, 1, 1, 12, 1, 1, 3, 3, 1, 1, 1, 5, 5, 1, 1, 3, 1, 1, 1
Offset: 1

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Author

Amiram Eldar, Oct 08 2022

Keywords

Comments

The corresponding sum of divisors of the powerful part of n is A295294.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, 1, e + 1]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(e = factor(n)[,2]); prod(i=1, #e, if(e[i] == 1, 1, e[i] + 1))};

Formula

a(n) = A000005(A057521(n)).
a(n) = A000005(n)/A056671(n).
a(n) = A000005(A064549(A003557(n))).
a(n) = 1 iff n is squarefree (A005117).
a(n) = A000005(n) iff n is powerful (A001694).
Multiplicative with a(p^e) = 1 if e = 1 and e+1 otherwise.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} ((p^3 - p^2 + 2*p - 1)/(p^2*(p - 1))) = 2.71098009471568319328... .
Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/p^s + 2/p^(2*s) - 1/p^(3*s)). - Amiram Eldar, Sep 09 2023

A383159 The sum of the maximum exponents in the prime factorizations of the unitary divisors of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 3, 2, 3, 1, 5, 1, 3, 3, 4, 1, 5, 1, 5, 3, 3, 1, 7, 2, 3, 3, 5, 1, 7, 1, 5, 3, 3, 3, 6, 1, 3, 3, 7, 1, 7, 1, 5, 5, 3, 1, 9, 2, 5, 3, 5, 1, 7, 3, 7, 3, 3, 1, 11, 1, 3, 5, 6, 3, 7, 1, 5, 3, 7, 1, 8, 1, 3, 5, 5, 3, 7, 1, 9, 4, 3, 1, 11, 3, 3, 3
Offset: 1

Views

Author

Amiram Eldar, Apr 18 2025

Keywords

Comments

First differs from A032741 at n = 36, and from A305611 and A325770 at n = 30.
a(n) depends only on the prime signature of n (A118914).

Examples

			4 has 2 unitary divisors: 1 and 4 = 2^2. The maximum exponents in their prime factorizations are 0 and 2, respectively. Therefore, a(4) = 0 + 2 = 2.
12 has 4 divisors: 1, 3 = 3^1, 4 = 2^2 and 12 = 2^2 * 3. The maximum exponents in their prime factorizations are 0, 1, 2 and 2, respectively. Therefore, a(12) = 0 + 1 + 2 + 2 = 5.
		

Crossrefs

Programs

  • Mathematica
    emax[n_] := If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]; a[n_] := DivisorSum[n, emax[#] &, CoprimeQ[#, n/#] &]; Array[a, 100]
    (* second program: *)
    a[n_] := If[n == 1, 0, Module[{e = FactorInteger[n][[;; , 2]], emax, v}, emax = Max[e]; v = Table[Times @@ (If[# < k + 1, 2, 1] & /@ e), {k, 1, emax}]; v[[1]] + Sum[k*(v[[k]] - v[[k - 1]]), {k, 2, emax}] - 1]]; Array[a, 100]
  • PARI
    emax(n) = if(n == 1, 0, vecmax(factor(n)[,2]));
    a(n) = sumdiv(n, d, (gcd(d, n/d) == 1) * emax(d));
    
  • PARI
    a(n) = if(n == 1, 0, my(e = factor(n)[, 2], emax = vecmax(e), v); v = vector(emax, k, vecprod(apply(x ->if(x < k+1, 2, 1), e))); v[1] + sum(k = 2, emax, k * (v[k]-v[k-1])) - 1);

Formula

a(n) = Sum_{d|n, gcd(d, n/d) = 1} A051903(d).
a(n) = A034444(n) * A383160(n)/A383161(n).
a(n) <= A383156(n), with equality if and only if n is squarefree (A005117).
a(n) = utau(n, 2) - 1 + Sum_{k=2..A051903(n)} k * (utau(n, k+1) - utau(n, k)), where utau(n, k) is the number of k-free unitary divisors of n (k-free numbers are numbers that are not divisible by a k-th power other than 1). For a given k >= 2, utau(n, k) is a multiplicative function with utau(p^e, k) = 2 if e < k, and 1 otherwise. E.g., utau(n, 2) = A056671(n), utau(n, 3) = A365498(n), and utau(n, 4) = A365499(n).
Sum_{k=1..n} a(k) ~ c_1 * n * log(n) + c_2 * n, where c_1 = c(2) + Sum_{k>=3} (k-1) * (c(k) - c(k-1)) = 0.91974850283445458744..., c(k) = Product_{p prime} (1 - 1/p^2 - 1/p^k + 1/p^(k+1)), c_2 = -1 + (2*gamma - 1)*c_1 + d(2) + Sum_{k>=3} (k-1) * (d(k) - d(k-1)) = -0.50780794945146599739..., d(k) = c(k) * Sum_{p prime} (2*p^(k-1) + k*p - k - 1) * log(p) / (p^(k+1) - p^(k-1) - p + 1), and gamma is Euler's constant (A001620).
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